F.4 maths

The y-intercept of the graph of the quadratic equation in two unknownsy= a(x-h)- (25/2) is 12. One of the x-intercept of the graph of theequation is 6.

a)
=> y= a(x-h)- (25/2) go though point (0,12) and (6,0)
=>12=a(0-h)^2-(25/2) --(1) and 0=a(12-h)^2 -25/2--(2)
=>49=2ah^2 --(1) and 25=2a(12-h)^2--(2)
(1)/(2):49/25=h^2/(12-h)^2
=>7056-1176+49h^2=25h^2
=>24h^2-1176h+7056=0
=>h^2-49h+294=0
=>(h-7)(h-42)=0
=>h=7 or h=42//
=>a=1/2 or 1/72//

b)
for h=7 , a=1/2 ,
we put y=0 ,
(1/2)(x-7)- (25/2)=0
h=12 or 2 ..

for h=42 and a=1/72,
we put y=0
(1/72)(x-42)- (25/2)=0
x=72 or x=12//


2008-12-28 15:48:31 補充：
myisland8132 :

你做少左負數個PART....

y= a(x-h)- (25/2) y-intercept 12, one x-intercept 6
So
12=ah^2-(25/2)...(1)
0=a(6-h)^2-(25/2)...(2)
And from (1)
a=49/2h^2
sub into (2)
(49/2h^2)(6-h)^2=(25/2)
[7(6-h)/h]=5
12h=42
h=7/2
a=2
(b)
y=2(x-7/2)- (25/2)=2x^2-14x+12=2(x^2-7x+6)=2(x-6)(x-1)
the other x-intercept of the graph is 1